Mathesis enucleata, or, The elements of the mathematicks by J. Christ. Sturmius ; made English by J.R. and R.S.S.

PROBLEM I.

TO find a square ABCD (such as in the mean while we'll suppose n. 1. to be in Fig. 41.) from which having taken away another square AEFG, which shall be half the former, ••here will be left the Rectangle GC whose Area is given. E. g. Suppose the given area equal to the square of the given line LM, to find the true sides of the squares AB and AE, answer∣••ng to these supposed ones, n. 1.

SOLƲTION.

Make the area of the rectangle that is to remain = bb, and GB=x; BC or AB will be = 〈 math 〉〈 math 〉, and substracting hence GB, the remaining side of the lesser square AG=〈 math 〉〈 math 〉−x, 〈◊〉〈◊〉. 〈 math 〉〈 math 〉. Since therefore the square of this is supposed ••o be half of the square of AB, this will be the Equation: 〈 math 〉〈 math 〉; ••nd multiplying by xx, 〈 math 〉〈 math 〉 ••nd multiplying by 2, 〈 math 〉〈 math 〉;

and substracting 2b{powerof4}, and adding 〈 math 〉〈 math 〉; and dividing by 2, 〈 math 〉〈 math 〉.

NB. The same Equation may be obtain'd, if, putting x for GB or FH, and having found the □ of AG or GF as a∣bove, you infer 〈 math 〉〈 math 〉.

This last Equation, tho' it be a biquadratick, yet may be rightly esteem'd only a quadratick one, because there is neither x{powerof3} nor single x in it, and so you may substitute this for it, 〈 math 〉〈 math 〉, viz. by supposing y=xx. Whence according to the third case of affected quadraticks, y will = 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 or = 〈 math 〉〈 math 〉.

Therefore 〈 math 〉〈 math 〉.

Geometrical Construction. Now if the given line b be assu∣med for unity, bb and b{powerof4} will be = = to the same line Therefore, if between LM as unity, and MN=½b viz 〈 math 〉〈 math 〉 you find a mean proportional MO (n. 2. Fig. 41.) that wil•• be = 〈 math 〉〈 math 〉, which being substracted from LM, and added to it, will give the two values of the quantity y. Moreover there∣fore by extracting its roots, i. e. by finding other mean pro∣portionals LR and LS between the quantities found LP and LQ and unity (n. 3.) they will be the two values of the quantity x sought; the first whereof LR will satisfie the que∣stion, and the other LS be impossible. Wherefore to form

the square it self, since its side will be = 〈 math 〉〈 math 〉; by making (n 4.) as x to b so b to a fourth, it will be obtain'd: And this may be further prov'd, if finding a mean proportional BK between BI=LR and the side of the □ BC, it be equal to the given quantity LM.

Arithmetical Rule. From the given area or the square of the given line LM substract the root of half the biquadrate of the same line; thus you will have the value of the □ FC, viz. xx: Therefore extracting further the square root of ••••is, it will be the value of x sought.